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Assume $W_{t}$ is a standard Brownian Motion, calculate the the probability that $W_{t}*W_{2t}$ is negative, i.e., $P(W_{t}*W_{2t}<0)$. I find it tricky to calculate the probability.Thank you.

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    $\begingroup$ Looks like homework. What have you tried? You probably want to use the fact that $W$ has independant increments. $\endgroup$ – AFK May 4 '15 at 12:01
  • $\begingroup$ This is conceptually simple, but you may need some tedious computations. $\endgroup$ – Gordon May 4 '15 at 12:53
  • $\begingroup$ I have tried to solve like: $P(W_{t}*W_{2t}<0)=P(W_{t}<0,W_{2t}>0)+P(W_{t}>0,W_{2t}<0)=$$P(W_{2t}>0|W_{t}<0)P(W_{t}<0)+P(W_{2t}<0|W_{t}>0)P(W_{t}>0)=P(W_{2t}-W_{t}+W_{t}>0|W_{t}<0)P(W_{t}<0)+P(W_{2t}-W_{t}+W_{t}<0|W_{t}>0)P(W_{t}>0)$. $\endgroup$ – cmd1991 May 4 '15 at 13:51
  • $\begingroup$ I have no idea how to continue, Is there any hint? Or are there any other solutions? $\endgroup$ – cmd1991 May 4 '15 at 13:53
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Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= P(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=E\Big(\mathbb{1}_{\{W_t/\sqrt{t} < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=\int_{-\infty}^0 \phi(x) \Phi(x) dx\\ &=\frac{1}{2}\Phi(x)^2\mid_{-\infty}^0\\ &=\frac{1}{8}, \end{align*} where \begin{align*} \phi(x)=\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \end{align*} is the density of a standard normal random variable, and $\Phi(x)$ is the cumulative distribution function.

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  • $\begingroup$ In the first line $P(W_t < 0, W_{2t} >0) = E(W_t < 0, W_{2t}-W_t > -W_t),$ why is probability changed to expectation? Which result are you using? $\endgroup$ – Idonknow Oct 30 '19 at 13:40
  • $\begingroup$ @Idonknow: Note that $P(A) = E(\pmb{1}_A)$. $\endgroup$ – Gordon Oct 30 '19 at 13:53
  • $\begingroup$ Yes, I am aware of this equality. But there is no indicator function in $E[W_t<0, W_{2t} - W_t > - W_t].$ $\endgroup$ – Idonknow Oct 30 '19 at 13:55
  • $\begingroup$ It is a typo. See the revision. $\endgroup$ – Gordon Oct 30 '19 at 13:56
  • $\begingroup$ I see. Thanks for clarifying. $\endgroup$ – Idonknow Oct 30 '19 at 13:57
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Since $W_{2t}-W_{t}$ is independent of $W_t$ and has the same law as $W_{2t-t}=W_t$ we only have to compute $$P(X(X+Y)<0)$$ where $(X,Y)$ follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either $X<0$ and $X+Y>0$ or the opposite. The two events have the same probability since $(-X,-Y)\sim (X,Y)$. You are left with the computation of $$ P(X<0,X+Y>0)$$ since the distribution of $(X,Y)$ is invariant by rotation around the z-axis) this probability can be computed geometrically (think cutting a cake, the cake being the bivariate density) : it is equal to $1/8$. The final result is thus $1/4$.

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  • $\begingroup$ Thank you very much for giving an intuitive answer. Just a bit confused that do you mean $(-X, -X-Y) \sim (X, X+Y)$ instead of $(-X, -X-Y) \sim (X, Y)$ $\endgroup$ – cmd1991 May 4 '15 at 15:53
  • $\begingroup$ That was a typo sorry. $\endgroup$ – vanna May 4 '15 at 18:02

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